Aditya Gupta
Last Activity: 5 Years ago
observe that point of intersection of the lines 2x + 3y =1 and x - y = 3 can found out by solving these eqns together, to obtain (2, – 1).
now, x^2 + y ^2 - 2x - 6y - 3= 0 can be written as (x – 1)^2 + (y – 3)^2= 13, so its centre is (1, 3).
as the other circle is conentric with this circle, it too will have the same centre. so its eqn is:
(x – 1)^2 + (y – 3)^2= r^2, where r is its radius. but, it passes through point of intersection of the lines 2x + 3y =1 and x - y = 3, which is (2, – 1).
so, (2 – 1)^2 + ( – 1 – 3)^2= r^2
or r^2= 17
hence eqn becomes
(x – 1)^2 + (y – 3)^2= 17
or x^2 + y^2 – 2x – 6y – 7= 0
kindly approve :))
